The electric field components in Fig. 1.27 are `E_(x)=alphax^(1//2), E_(y)=E_(z)=0,` in which `alpha=800 N//C m^(1//2)`. Calculate (a) the flux through the cube, and (b) the charge withing the cube. Assume that `alpha=0.1m`.

The electric field components in Figure are E_(x)=alpha x^(1/2), E_y=E_(z)=0 in which alpha=800 N//Cm^(1/2) . Calculate (a) the flux through the cube, and (b) the charge within the cube. Assume that a=0.1m.

The electric field components in Fig. are E_(x)=ax^(1//2), E_(y)=E_(z)=0 , in which alpha=800N//C m^(1//2) . Calculate the flux through the cube

The electric field components are E_(x)=ax^(-1//2),E_(y)=E_(z)=0 in which a =800 N/C m^(1//2) calculate (a) the flux through the cube and (b) the charge within the cube asume that a=0.1 m

The electric field components are E_(x)=ax^(-1//2),E_(y)=E_(z)=0 in which a =800 N/C m^(1//2) calculate (a) the flux through the cube and (b) the charge within the cube asume that a=0.1 m

The electric field components are E_(x)=ax^(1//2),E_(y)=E_(z)=0 in which a=800NC^(-1)m^(-1//2) . Calculate a. the flux through the cube, and b. the charge within the cube. Assume that a=0.1m

Using Gauss’s law, derive expression for intensity of electric field at any point near the infinitely long straight uniformly charged wire. The electric field components in the following figure are E_(x) = alphax, E_(y) = 0, E_(z) = 0, " in which " alpha = 400 N//C m. Calculate (i) the electric flux through the cube, and (ii) the charge within the cube assume that a = 0.1m.

Using Gauss’s law, derive expression for intensity of electric field at any point near the infinitely long straight uniformly charged wire. The electric field components in the following figure are E_(x) = alphax, E_(y) = 0, E_(z) = 0, " in which " alpha = 400 N//C m. Calculate (i) the electric flux through the cube, and (ii) the charge within the cube assume that a = 0.1m.